工程电磁场第八版后答案第04章.pdfVIP

CHAPTER 4 4.1. The value of E at P (ρ = 2, φ = 40◦ , z = 3) is given as E = 100aρ − 200aφ + 300az V/m. Determine the incremental work required to move a 20 µC charge a distance of 6 µm: a) in the direction of aρ : The incremental work is given by dW = −q E · dL, where in this case, dL = dρ aρ = 6 × 10−6 aρ . Thus dW = −(20 × 10−6 C)(100 V/m)(6 × 10−6 m) = −12 × 10−9 J = −12 nJ b) in the direction of aφ : In this case dL = 2 dφ aφ = 6 × 10−6 aφ , and so dW = −(20 × 10−6 )(−200)(6 × 10−6 ) = 2.4 × 10−8 J = 24 nJ c) in the direction of az : Here, dL = dz az = 6 × 10−6 az , and so dW = −(20 × 10−6 )(300)(6 × 10−6 ) = −3.6 × 10−8 J = −36 nJ d) in the direction of E: Here, dL = 6 × 10−6 aE , where 100aρ − 200aφ + 300az aE = 2 2 2 1/2 = 0.267 aρ − 0.535 aφ + 0.802 az [100 + 200 + 300 ] Thus dW = −(20 × 10−6 )[100a − 200a + 300a ] · [0.267 a − 0.535 a + 0.802 a ](6 × 10−6 ) ρ φ z ρ φ z = −44.9 nJ e) In the direction of G = 2 ax − 3 ay + 4 az : In this case, dL = 6 × 10−6 aG , where 2ax − 3ay + 4az aG = 2 2 2 1/2 = 0.371 ax − 0.557 ay + 0.743 az [2 + 3 + 4 ] So now dW = −(20 × 10−6 )[100a − 200a + 300a ] · [0.371 a − 0.557 a + 0.743 a ](6 × 10−6 ) ρ φ z x y z = −(20 × 10−6 ) [37.1(aρ · ax ) − 55.7(aρ · ay ) − 74.2(aφ · ax ) + 111.4(aφ · ay )